\(-26-\left(x-7\right)=0\\ \Rightarrow x-7=-26\\ \Rightarrow x=-19\)

\(\Rightarrow-26-x+7=0\\ \Rightarrow x+19=0\Rightarrow x=-19\)

@ Bắp gửi tus ạ

\(-26 - (x - 7) = 0 \)

\(=> x - 7 = -26\)

\(=> x = -19\)

không biết

-26-(x-7)=0

-26-x+7=0

-21-x=0

x=-21-0

x=-21

Vậy x=-21

tk nha!

-26-(x-7)=0 \(\Rightarrow\chi-7=-26\)

\(\Rightarrow\chi=-26+7\)

\(\Rightarrow\chi=-19\)

HTDT

-26-(x-7)=0

-26-x+7=0

-21-x=0

x=-21-0

x=-21

Cách 1 :

26 - ( x - 7 ) = 0

( x - 7 ) = 26 - 0

x - 7 = 26

x = 26 + 7

x = 33

Cách 2 :

26 - ( x - 7 ) = 0

26 - x - 7 = 0 

26 + 7 - x = 0

33 - x = 0

x = 33 - 0

x = 33

26 - | x-7 | =0

|x-7| = 26 

Th1: x-7 = -26

x= -19

Th2: x-7 = 26 

x= 33

Vậy x= -19 và x= 33

C2 : ko bt.

Cách 1 :

26 - [ x - 7 ] = 0

=> 26 - x + 7 = 0

=> 33 - x = 0

=> x = 33

Cách 2

26 - [x - 7] = 0

=> x - 7 =26

=> x = 33

toán lp 1 ???

toán lớp 1 sao học ghê vậy lm đc cả x vs ^ luôn ô mai gi gứ chóp bạn nào lớp 1 mà giải đc bài này luôn ?????

\(a,3-\left(17-x\right)=-12\\ \Rightarrow17-x=15\\ \Rightarrow x=2\\ b,-26-\left(x-7\right)=0\\ \Rightarrow x-7=-26\\ \Rightarrow x=-19\\ c,25+\left(-2+x\right)=5\\ \Rightarrow-2+x=20\\ \Rightarrow x=18\\ d,30+\left(32-x\right)=10\\ \Rightarrow32-x=-20\\ \Rightarrow x=52\)

a) `3-(17-x)=-12`

`3-17+x=-12`

`x=-12-3+17`

`x=2`

b) `-26-(x-7)=0`

`-26-x+7=0`

`-19-x=0`

`x=-19`

c) `25+(-2+x)=5`

`25-2+x=5`

`x=5-25+2`

`x=-18`

d) `30+(32-x)=10`

`30+32-x=10`

`62-x=10`

`x=52`

a/ \(x\left(x+7\right)=0\) \(\Rightarrow\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

b/ \(\left(-x+5\right)\left(3-x\right)=0\) \(\Rightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)

c/ \(\left|x-1\right|=3\) \(\Rightarrow\left[\begin{matrix}x-1=3\\1-x=3\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

d/ \(-13\left|x\right|=-26\) \(\Rightarrow\left|x\right|=2\) \(\Rightarrow x=\pm2\)

e/ \(x.x-8=-2.\left(-13\right)-\left(-2\right)\)

\(\Rightarrow x^2=36\) \(\Rightarrow\left|x\right|=6\) \(\Rightarrow x=\pm6\)

*) \(-7+2x=-37-\left(-26\right)\)

⇔ \(-7+2x=-11\)

⇔ \(2x=\left(-11\right)-\left(-7\right)\)

⇔ \(2x=-4\)

⇔ \(x=\left(-4\right):2\)

⇔ \(x=-2\)

\(\text{(3x +9).(11-x ) =0}\)

⇔\(\left[{}\begin{matrix}3x+9=0\\11=x=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}3x=-9\\x=11\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=-3\\x=11\end{matrix}\right.\)

Vậy x= -3 và x= 11

⇔⇔⇔

Vậy x= -3 và x= 11